將上傳的文件保存在特定位置

Question

我有以下處理文件在服務器上上傳的代碼。但如何將文件保存到特定位置的服務器上

import gwtupload.server.UploadAction; 
import gwtupload.server.exceptions.UploadActionException; 

import org.apache.commons.fileupload.FileItem; 

import java.io.File; 
import java.io.FileInputStream; 
import java.io.IOException; 
import java.util.Hashtable; 
import java.util.List; 

import javax.servlet.http.HttpServletRequest; 
import javax.servlet.http.HttpServletResponse; 




/** 
* This is an example of how to use UploadAction class. 
* 
* This servlet saves all received files in a temporary folder, 
* and deletes them when the user sends a remove request. 
* 
* @author Manolo Carrasco Moñino 
* 
*/ 
public class SampleUploadServlet extends UploadAction { 

    private static final long serialVersionUID = 1L; 

    Hashtable<String, String> receivedContentTypes = new Hashtable<String, String>(); 
    /** 
    * Maintain a list with received files and their content types. 
    */ 
    Hashtable<String, File> receivedFiles = new Hashtable<String, File>(); 

    /** 
    * Override executeAction to save the received files in a custom place 
    * and delete this items from session. 
    */ 
    @Override 
    public String executeAction(HttpServletRequest request, List<FileItem> sessionFiles) throws UploadActionException { 
    String response = ""; 
    int cont = 0; 
    for (FileItem item : sessionFiles) { 
     if (false == item.isFormField()) { 
     cont++; 
     try { 
      /// Create a new file based on the remote file name in the client 
      // String saveName = item.getName().replaceAll("[\\\\/><\\|\\s\"'{}()\\[\\]]+", "_"); 
      // File file =new File("/tmp/" + saveName); 

      /// Create a temporary file placed in /tmp (only works in unix) 
      // File file = File.createTempFile("upload-", ".bin", new File("/tmp")); 

      /// Create a temporary file placed in the default system temp folder 
      File file = File.createTempFile("upload-", ".bin"); 
      item.write(file); 

      /// Save a list with the received files 
      receivedFiles.put(item.getFieldName(), file); 
      receivedContentTypes.put(item.getFieldName(), item.getContentType()); 

      /// Compose a xml message with the full file information 
      response += "<file-" + cont + "-field>" + item.getFieldName() + "</file-" + cont + "-field>\n"; 
      response += "<file-" + cont + "-name>" + item.getName() + "</file-" + cont + "-name>\n"; 
      response += "<file-" + cont + "-size>" + item.getSize() + "</file-" + cont + "-size>\n"; 
      response += "<file-" + cont + "-type>" + item.getContentType() + "</file-" + cont + "type>\n"; 
     } 
     catch (Exception e) 
     { 
     } 
     } 
    } 

    /// Remove files from session because we have a copy of them 
    removeSessionFileItems(request); 

    /// Send information of the received files to the client. 
    return "<response>\n" + response + "</response>\n"; 
    } 

    /** 
    * Get the content of an uploaded file. 
    */ 
    @Override 
    public void getUploadedFile(HttpServletRequest request, HttpServletResponse response) throws IOException { 
    String fieldName = request.getParameter(PARAM_SHOW); 
    File f = receivedFiles.get(fieldName); 
    if (f != null) { 
     response.setContentType(receivedContentTypes.get(fieldName)); 
     FileInputStream is = new FileInputStream(f); 
     copyFromInputStreamToOutputStream(is, response.getOutputStream()); 
    } else { 
     renderXmlResponse(request, response, ERROR_ITEM_NOT_FOUND); 
    } 
    } 

    /** 
    * Remove a file when the user sends a delete request. 
    */ 
    @Override 
    public void removeItem(HttpServletRequest request, String fieldName) throws UploadActionException { 
    File file = receivedFiles.get(fieldName); 
    receivedFiles.remove(fieldName); 
    receivedContentTypes.remove(fieldName); 
    if (file != null) { 
     file.delete(); 
    } 
    } 

} 

Answer 1

您應該使用File#createTempFile() which takes a directory來代替。

File file = File.createTempFile("upload-", ".bin", new File("/path/to/your/uploads")); 
item.write(file); 

或者，如果你真的想臨時文件移動到另一個位置事後，使用File#renameTo()。

File destination = new File("/path/to/your/uploads", file.getName()); 
file.renameTo(destination); 

Answer 2

在您的java應用程序中，您可以創建具有特定路徑的文件。

贊; 「新的文件（」 C：/文件/文件名）」。等

見Java Documents瞭解新IO包的文件類的CreateFile函數